/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 A \(12.6-\mathrm{V}\) car batter... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(12.6-\mathrm{V}\) car battery with negligible internal resistance is connected to a series combination of a \(3.2-\Omega\) resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law but instead has a current-voltage relationship \(V=\alpha I+\beta I^{2},\) with \(\alpha=\) 3.8\(\Omega\) and \(\beta=1.3 \Omega / \mathrm{A} .\) What is the current through the \(3.2-\Omega\) resistor?

Short Answer

Expert verified
The current through the 3.2-Ω resistor is approximately 1.40 A.

Step by step solution

01

Express total voltage across components

The total voltage across the series combination of the resistor and the thermistor is equal to the voltage of the battery, which is 12.6 V.
02

Determine voltage across the 3.2-Ω resistor

According to Ohm's law, the voltage across the resistor can be written as \( V_R = I \cdot R \). For a 3.2 Ω resistor, this is \( V_R = I \cdot 3.2 \).
03

Formulate the equation for the thermistor

The given relationship for the thermistor is \( V_T = \alpha I + \beta I^2 \). Substituting the given values, \( \alpha = 3.8 \) Ω and \( \beta = 1.3 \) Ω/A, we get \( V_T = 3.8I + 1.3I^2 \).
04

Apply Kirchhoff’s voltage law to the circuit

For the entire circuit, the sum of the voltage drops equals the battery voltage: \( V_R + V_T = 12.6 \). Substituting the expressions we have: \( 3.2I + 3.8I + 1.3I^2 = 12.6 \).
05

Simplify and solve the quadratic equation

Combine the linear terms for \(I\) to simplify the equation: \( 7I + 1.3I^2 = 12.6 \). Rearrange this into a standard quadratic form: \( 1.3I^2 + 7I - 12.6 = 0 \). Use the quadratic formula \( I = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1.3, b=7, c=-12.6 \).
06

Calculate the roots using the quadratic formula

Calculate the discriminant \( b^2 - 4ac \):\(7^2 - 4 \cdot 1.3 \cdot (-12.6) = 49 + 65.52 = 114.52\).Solve for \(I\):\( I = \frac{-7 \pm \sqrt{114.52}}{2 \cdot 1.3} \).Calculate the two possible values for \(I\):\( I_1 = \frac{-7 + \sqrt{114.52}}{2.6} \) and \( I_2 = \frac{-7 - \sqrt{114.52}}{2.6} \).Since current cannot be negative, we focus on \( I_1 \).
07

Conclusion of current value

By calculating \( I_1 \) precisely, find the positive current value that suits the physical context of the circuit: \( I \approx 1.40 \) A.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kirchhoff’s Voltage Law
In the world of electrical engineering, Kirchhoff's Voltage Law (KVL) is a fundamental principle used in circuit analysis. It states that the sum of the electrical voltage around any closed network is zero. This means that the energy provided by power sources is completely accounted for by voltage drops across circuit elements.

To understand KVL, imagine a simple loop in a circuit containing a battery and a few components like resistors and thermistors in series. The battery introduces a voltage rise, and as current moves through resistors and other elements, voltages drop across these components. According to Kirchhoff's Voltage Law, the sum of these rises and drops equals zero. Essentially, in a circuit loop, the total voltage provided by sources must be entirely converted to potential difference across the load elements.

This law is crucial when analyzing complex circuits as it helps in writing equations that describe the relations of voltage across different components. It ensures consistency in the conservation of energy principle—a core reason why circuits function as intended.
Quadratic Equation
In solving the problem involving a thermistor that does not obey Ohm’s Law directly, we encounter a quadratic equation. A quadratic equation is of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and it can have two solutions.

In our specific exercise, we arrive at a quadratic equation after applying Kirchhoff's Voltage Law to the circuit. We have variables of current \( I \) squared term due to the thermistor's behavior being defined as \( V_T = \alpha I + \beta I^2 \). We then perform a common practice of using the quadratic formula \( I = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

The quadratic formula is a powerful tool in mathematics to find unknowns in equations of this type. Here, \( a = 1.3 \), \( b = 7 \), and \( c = -12.6 \). Calculating the discriminant \( b^2 - 4ac \) helps determine the nature of solutions. If the result is positive, two real solutions exist—the scenario we observe leading to a viable \( I \) for our circuit.
Circuit Analysis
Circuit analysis involves the process of finding current and voltage across each element of an electrical circuit. It integrates laws like Kirchhoff's along with mathematical tools such as solving linear and non-linear equations.

The essence of this technique helps to predict how circuits function under different conditions. In our exercise, the analysis revolved around calculating the current through the resistor combined in series with a thermistor. Circuit analysis begins by identifying all known components and their relationships like resistive values and voltage sources.

Drawing circuit diagrams helps visualize connections. Then apply systematic approaches such as writing KVL for voltage loops and using Ohm's Law to determine current through resistors. In circuits involving non-linear components like our thermistor, specialized relationships \( V = \alpha I + \beta I^2 \) are considered which leads to solving equations to find desired quantities like current or voltage.
  • Identify known and unknown parameters.
  • Set up equations using core principles like KVL and Ohm's Law.
  • Simplify and solve using appropriate mathematical approaches.
Mastering these steps allows not only solving academic problems but also real-world circuit design and diagnostics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An idealized voltmeter is connected across the terminals of a \(15.0-\mathrm{V}\) battery, and a \(75.0-\Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{V} :\) (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

A silver wire 2.6 \(\mathrm{mm}\) in diameter transfers a charge of 420 \(\mathrm{C}\) in 80 \(\mathrm{min}\) . Silver contains \(5.8 \times 10^{28}\) free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. A 71.0 -cm length of 12 -gauge copper wire that is 2.05 \(\mathrm{mm}\) in diameter carries 4.85 A of current. (a) How much time does it take for an electron to travel the length of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 \(\mathrm{mm}\) ) of the same length that carries the same current.(c) Generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

A \(25.0-\Omega\) bulb is connected across the terminals of a \(12.0-\mathrm{V}\) battery having 3.50\(\Omega\) of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

A source with emf \(\mathcal{E}\) and internal resistance \(r\) is connected to an external circuit. (a) Show that the power output of the source is maximum when the current in the circuit is one-half the short-circuit current of the source. (b) If the external circuit consists of a resistance \(R,\) show that the power output is maximum when \(R=r\) and that the maximum power is \(\mathcal{E}^{2} / 4 r_{1}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.